A beam AB of mass m and of uniform cross section is suspended from two springs as shown. If spring 2 breaks, determine at that instant (a) the angular acceleration of the bar, (b) the acceleration of Point A, (c) the acceleration of Point B.

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Statics:[latex]{ T }_{ 1 }={ T }_{ 2 }=\frac { 1 }{ 2 } W=\frac { 1 }{ 2 } mg[/latex]

(a) Angular acceleration: [latex] +\curvearrowright \sum { { M }_{ G } } =\sum { { \left( { M }_{ G } \right) }_{ eff } } :T\left( \frac { L }{ 2 } \right) =\overline { I } \alpha [/latex]

[latex]\ \frac { 1 }{ 2 } mg\left( \frac { L }{ 2 } \right) =\frac { 1 }{ 12 } m{ L }^{ 2 }\alpha [/latex]

[latex]\ \alpha =\frac { 3g }{ L } \quad \quad ,\alpha =\frac { 3g }{ L } \curvearrowright [/latex]

[latex]+\downarrow \sum { { F }_{ y } } =\sum { { \left( { F }_{ Y } \right) }_{ eff } } :W-{ T }_{ 1 }=m\overline { a } [/latex]

[latex]\ mg-\frac { 1 }{ 2 } mg=m\overline { a } [/latex]

[latex]\ \overline { a } =\frac { 1 }{ 2 } g\quad \quad \overline { a } =\frac { 1 }{ 2 } g\downarrow [/latex]

(b) Acceleration of A: [latex]{ a }_{ A }={ a }_{ G }+{ a }_{ A/G }[/latex]

[latex]\ +\downarrow { a }_{ A }=\frac { 1 }{ 2 } g-\frac { L }{ 2 } \alpha[/latex]

[latex]\ =\frac { 1 }{ 2 } g-\frac { L }{ 2 } \left( \frac { 3g }{ L } \right) [/latex]

[latex]\ { a }_{ A }=-g\quad ,{ a }_{ A }=g\uparrow [/latex]

(c) Acceleration of B: [latex]{ a }_{ B }={ a }_{ G }+{ a }_{ B/G }[/latex]

[latex]\ +\downarrow { a }_{ B }=\overline { a } +\frac { 1 }{ 2 } \alpha =\frac { 1 }{ 2 } g+\frac { L }{ 2 } \left( \frac { 3g }{ L } \right) =+2g\[/latex]

[latex]\ \quad { a }_{ B }=2g\downarrow [/latex]

Question : A beam AB of mass m and of uniform cross section ...

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